Why Multiple file uploads in FastAPI? - Purpose & Use Cases

The Big Idea

Discover how to handle many files at once without messy code!

The Scenario

Imagine you want users to send several photos at once through a web form, but you have to handle each file separately in your code.

The Problem

Manually processing each file one by one means writing repetitive code, increasing chances of mistakes, and making your app slower and harder to maintain.

The Solution

FastAPI lets you accept multiple files easily with a simple parameter, automatically handling the uploads and letting you focus on what to do with the files.

Before vs After

✗ Before

async def upload(file1: UploadFile, file2: UploadFile):
    content1 = await file1.read()
    content2 = await file2.read()
    # process files separately

✓ After

async def upload(files: List[UploadFile]):
    for file in files:
        content = await file.read()
        # process each file in a loop

What It Enables

You can build smooth user experiences that accept many files at once without complicated code.

Real Life Example

Think of a photo-sharing app where users upload dozens of pictures in one go, and your backend handles them all effortlessly.

Key Takeaways

Manual file handling is repetitive and error-prone.

FastAPI simplifies multiple file uploads with a single parameter.

This makes your code cleaner and your app more user-friendly.

Practice

1. In FastAPI, which type hint should you use to accept multiple uploaded files in an endpoint?

easy

A. str with UploadFile

B. List[UploadFile] with File(...)

C. List[str] with File(...)

D. UploadFile without List

Solution

1. Step 1: Understand FastAPI file upload types

   FastAPI uses UploadFile to handle file uploads efficiently.

2. Step 2: Use List for multiple files

   To accept multiple files, you wrap UploadFile in a List and use File(...) to declare it as a file input.

3. Final Answer:

   List[UploadFile] with File(...) -> Option B

4. Quick Check:

   Multiple files = List[UploadFile] [OK]

Hint: Use List[UploadFile] with File(...) for multiple files [OK]

Common Mistakes:

• Using UploadFile without List for multiple files
• Using List[str] which expects strings, not files
• Confusing str type with file uploads

2. Which of the following is the correct syntax to declare a FastAPI endpoint that accepts multiple files named 'files'?

easy

A. files: List[UploadFile] = File(...)

B. files: UploadFile = File(...)

C. files: List[str] = File(...)

D. files: str = UploadFile(...)

Solution

1. Step 1: Check parameter type for multiple files

   Multiple files require List[UploadFile] type hint.

2. Step 2: Use File(...) to mark as file input

   File(...) is needed to tell FastAPI this is a file upload field.

3. Final Answer:

   files: List[UploadFile] = File(...) -> Option A

4. Quick Check:

   Multiple files syntax = List[UploadFile] = File(...) [OK]

Hint: Use List[UploadFile] = File(...) for multiple files param [OK]

Common Mistakes:

• Using UploadFile without List for multiple files
• Using List[str] which is incorrect for files
• Assigning UploadFile(...) instead of File(...)

3. Given this FastAPI endpoint code, what will be the output if two files named 'a.txt' and 'b.txt' are uploaded?

from fastapi import FastAPI, File, UploadFile
from typing import List

app = FastAPI()

@app.post('/upload')
async def upload(files: List[UploadFile] = File(...)):
    return {"filenames": [file.filename for file in files]}

medium

A. {"filenames": ["a.txt", "b.txt"]}

B. {"filenames": ["files", "files"]}

C. {"filenames": []}

D. Runtime error due to wrong type

Solution

1. Step 1: Understand the endpoint logic

   The endpoint returns a dictionary with a list of filenames extracted from the uploaded files.

2. Step 2: Check the uploaded files names

   Two files named 'a.txt' and 'b.txt' are uploaded, so their filenames will be returned in the list.

3. Final Answer:

   {"filenames": ["a.txt", "b.txt"]} -> Option A

4. Quick Check:

   Returned filenames list matches uploaded files [OK]

Hint: Returned filenames list matches uploaded files names [OK]

Common Mistakes:

• Expecting file content instead of filenames
• Confusing parameter name with file names
• Assuming empty list if files uploaded

4. What is wrong with this FastAPI endpoint for multiple file uploads?

from fastapi import FastAPI, File, UploadFile
from typing import List

app = FastAPI()

@app.post('/upload')
async def upload(files: UploadFile = File(...)):
    return {"count": len(files)}

medium

A. File(...) should be replaced with UploadFile(...)

B. Missing async keyword in function definition

C. len(files) is invalid because files is a string

D. files should be List[UploadFile] to accept multiple files

Solution

1. Step 1: Check parameter type for multiple files

   The parameter 'files' is typed as UploadFile, which accepts only one file.

2. Step 2: Correct type for multiple files

   To accept multiple files, it must be List[UploadFile].

3. Final Answer:

   files should be List[UploadFile] to accept multiple files -> Option D

4. Quick Check:

   Multiple files need List[UploadFile] type [OK]

Hint: Use List[UploadFile] for multiple files, not UploadFile alone [OK]

Common Mistakes:

• Using UploadFile instead of List[UploadFile] for multiple files
• Confusing File(...) with UploadFile(...)
• Forgetting async keyword (though present here)

5. You want to create a FastAPI endpoint that accepts multiple files and returns a JSON with each file's name and size in bytes. Which code snippet correctly implements this?

hard

A. async def upload(files: List[str] = File(...)):
return [{"name": f, "size": 0} for f in files]

B. async def upload(files: UploadFile = File(...)):
return [{"name": files.filename, "size": len(await files.read())}]

C. async def upload(files: List[UploadFile] = File(...)):
results = []
for file in files:
content = await file.read()
results.append({"name": file.filename, "size": len(content)})
return results

D. async def upload(files: List[UploadFile] = File(...)):
return [{"name": file.name, "size": file.size} for file in files]

Solution

1. Step 1: Accept multiple files with correct type

   Use List[UploadFile] with File(...) to accept multiple files.

2. Step 2: Read each file content to get size

   Use await file.read() to get bytes, then len() to get size.

3. Step 3: Return list of dicts with name and size

   Build a list of dictionaries with filename and size for each file.

4. Final Answer:

   Code snippet that correctly reads files and returns name and size -> Option C

5. Quick Check:

   Read files async, return name and size list [OK]

Hint: Read files async, use len(content) for size, return list [OK]

Common Mistakes:

• Using UploadFile without List for multiple files
• Trying to get size without reading file content
• Using wrong attribute names like file.name or file.size